2 Answers
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How is that integral infinite? If $x\neq 0$, then the function is locally integrable near $y=x$ and locally integrable near $y=0$ and it is integrable at $\infty$ since the integrand behaves like $|y|^{-2+\alpha+\beta}$ which decays like $y^{1+\epsilon}$. So the integral is not infinite unless $x=0$.

There is also the issue that the integral clearly depends on $x$, while the right side does not. I believe the typo is that $x$ should be 1.

If one carefully writes out the identity $I_\alpha(I_\beta(f))(z)=I_{\alpha+\beta}(f)(z)$ and sets $z=0$, one finds that

Thanks Peter Luthy. I changed my typos where $x$ should be $1$ in the one dimensional case since I integrate from 0 to 1. And I also changed $\Gamma$ to $\gamma$ since they are different. I mistook for the same function. Your proof of Setin's result is very nice. In his book, he just mentions that this identity is a consequence of the semigroup property that you use. You fill up the missing step. Thanks.
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AnandAug 10 '11 at 7:41

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You're welcome. I can totally understand the confusion about the domain of integration since Stein mentions the beta integral which only integrates on [0,1] and makes a typo that $x=1$! Then again, one often learns more from correcting mistakes in books than from simply reading. There is something to be said for getting one's hands dirty and doing some hard work.
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Peter LuthyAug 10 '11 at 8:49